Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $y = \dfrac{7t + 7}{t^2 - 10t + 16} \div \dfrac{t + 1}{6t - 12} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{7t + 7}{t^2 - 10t + 16} \times \dfrac{6t - 12}{t + 1} $ First factor the quadratic. $y = \dfrac{7t + 7}{(t - 2)(t - 8)} \times \dfrac{6t - 12}{t + 1} $ Then factor out any other terms. $y = \dfrac{7(t + 1)}{(t - 2)(t - 8)} \times \dfrac{6(t - 2)}{t + 1} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ 7(t + 1) \times 6(t - 2) } { (t - 2)(t - 8) \times (t + 1) } $ $y = \dfrac{ 42(t + 1)(t - 2)}{ (t - 2)(t - 8)(t + 1)} $ Notice that $(t + 1)$ and $(t - 2)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ 42(t + 1)\cancel{(t - 2)}}{ \cancel{(t - 2)}(t - 8)(t + 1)} $ We are dividing by $t - 2$ , so $t - 2 \neq 0$ Therefore, $t \neq 2$ $y = \dfrac{ 42\cancel{(t + 1)}\cancel{(t - 2)}}{ \cancel{(t - 2)}(t - 8)\cancel{(t + 1)}} $ We are dividing by $t + 1$ , so $t + 1 \neq 0$ Therefore, $t \neq -1$ $y = \dfrac{42}{t - 8} ; \space t \neq 2 ; \space t \neq -1 $